A chemistry course to cover selected topics covered in advanced high school chemistry courses, correlating to the standard topics as established by the American Chemical Society.
Prerequisites: Students should have a background in basic chemistry including nomenclature, reactions, stoichiometry, molarity and thermochemistry.

Reviews

JP

Excellent course. If you are struggling with high school chem, or need a good foundation for college chem - this course moves you through the concepts and gives you lots of practice problems.

PR

May 26, 2020

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Excellent Course. I sincerely suggest to introduce free courses like these in Computational Chemistry and Analytical Chemistry too that are highly useful to Graduate Chemistry Students

From the lesson

Chemical Equilibrium

This unit introduces the concept of chemical equilibrium and how it applies to many chemical reactions. The quantitative aspects of equilibrium are explored thoroughly through discussions of the law of mass action as well as the relationship between equilibrium constants with respect to concentrations and pressures of substances. Much of the discussion explores how to solve problems to find either the value of the equilibrium constant or the concentrations of substances at equilibrium. ICE (initial-change-equilibrium) tables are introduced as a problem-solving tool and multiple examples of their use are included.

From a qualitative standpoint, Le Châtelier’s principle is used to explain how various factors affect the equilibrium constant of a reaction along with the concentrations of all species.

Taught By

Dr. Allison Soult

Dr. Kim Woodrum

Transcript

In this module, we will explore Le Chatelier's principle and how it effects the system at chemical equilibrium. Because there are many examples to look at we will divide this into two parts. Because there are many examples to look at we will divide this into two parts. Our objective in this unit is to understand the concept of Le Chatelier's principle and to be able to predict how different changes will effect the system at chemical equilibrium. Before we can start looking at the specific details we need to understand the general concept of Le Chatelier's principle. What it tells us is, that when a chemical system, or chemical reaction What it tells us is, that when a chemical system, or chemical reaction is at equilibrium is disturbed the system shifts in a direction to minimize the disturbance. Well we see this in our everyday lives if you have ever been on an elevator. If there is one person on a elevator and second person gets on everyone rearranges in order to spread themselves out. A third person gets on and we all move again a fourth, and a fifth until we get to the point where we just cannot move anymore. but we try to spread ourselves out in the elevator. We rarely ever see an elevator door open and find four people standing really close together on one side of the elevator, and the other side being empty. As each person get on the elevator we are disturbing the equilibrium and the system, which is the people already in the elevator have to respond, or shift and the system, which is the people already in the elevator have to respond, or shift in order to minimize that disturbance. So how to we related Le Chatelier's principle back to the concepts we already have explored with equilibrium? Well if we look at a room full of people. We see that all the people are in the room now this systems is not at equilibrium We see that all the people are in the room now this systems is not at equilibrium it has got to go to equilibrium. now this systems is not at equilibrium it has got to go to equilibrium. And so to get to equilibrium what we see, is that 3 people leave the room and when those 3 people are outside of the room our system is now at equilibrium. Remember equilibrium is products over reactants. Or in this case we can look at it as the ratio of the number of people outside the room compared to the number of people inside the room so here we have a system at equilibrium we have 3 people outside the room and rest of the people inside the room. At time goes on we have different people outside the room here our blue people go back into the room and the purple people have left. We are still at equilibrium. Because what we are seeing is the number of people outside the room is still 3 and the rest of the people are inside the room and so our ratio remains the same. Now we have a system at equilibrium again and we disturb that equilibrium in some way. We are not going to worry about what that disturbance is. Say we add some more people. We are not going to worry about what that disturbance is. Say we add some more people. We change something about the ratio when we disturb that system at equilibrium. We change something about the ratio when we disturb that system at equilibrium. What we see is that a new equilibrium is established. So now this new equilibrium has 4 people outside of the room and the rest of the people inside. But the system had to respond to that disturbance to the equilibrium. Maybe we turned up the heat, maybe the music changed maybe they ran out of food, something changed To disturb that system which was at equilibrium and it had to respond and at this point equilibrium was restored. and it had to respond and at this point equilibrium was restored. There are many things that can effect a system at equilibrium. All of these disturbances can happen to reaction or to some system. And the system will respond in such a way to minimize this disturbance. So we can look at a variety of topics and we will actually go through examples of each of these, so we can see how they apply to actual chemical reaction. We can change the concentration of our reactant or product. We can change the pressure of the system and we can change the pressure in two different ways. One by changing the volume of the container or the addition of an inert gas. Remember that inert just means nonreactive. And what we will find is that that inert gas actually does not have any effect on the equilibrium. We can also change the temperature and we have to worry about is this an exothermic process, where heat is being released or is this an endothermic process where heat is being consumed. Because changes in temperature will affect those reactions in different ways. Because changes in temperature will affect those reactions in different ways. To lets look at an example here where we are looking at a reaction, we have our balanced chemical equation Carbon Monoxide reacting with Hydrogen to produce Methanol. And what we are going to look at is we are going to change the concentration of a reactant. And so what happens if we increase the concentration of a reactant. We are going to see the formation of more product. It is going to favor the formation of products. Because if I add additional carbon monoxide for example. Now I have disturbed my equilibrium we know that k equals concentrations of our products over the concentration of our reactants. And if I add reactants I now have a bigger number on the problem That I would need to have in order to balance and have the correct ratio with the concentration with the products. What needs to happen is that some of the excess CO that what added needs to be consumed and formed into products. As a result, we will also a reduction in the amount of H_2 present because it is needed to react with the CO. reduction in the amount of H_2 present because it is needed to react with the CO. We are going to see and increase in the amount of CH_3OH formed because we need to form products we need to increase the concentration of products and we need to decrease concentration of reactants which we have also done. And what this will allow the system to do is to return to equilibrium. Where we maintain that constant value for the equilibrium constant. to return to equilibrium. Where we maintain that constant value for the equilibrium constant. So now what happens if I decrease the amount of a reactant? Lets say I remove some carbon monoxide. I am not worried about the logistics or the experimental parameters or the setup I need to remove it. I am not worried about the logistics or the experimental parameters or the setup I need to remove it. but the amount of CO available decreases. What happens is the concentration of our reactants decrease and now we have disturbed this ratio. We are no longer a system at equilibrium. In order to restore the system to equilibrium it is going to have to compensate for that lost carbon monoxide and as a result the reaction for that lost carbon monoxide and as a result the reaction is going to shift to the left for favor the formation of more reactants. Some of our CH_3OH will decompose. We will increase the amount of H_2 a little bit. We will increase the amount of CO a little bit. Until we get to the point where the concentration of the products over the concentration of our reactants is equal to k. Now at this point, we are going to have a different amount of CO a different amount of H_2 Now at this point, we are going to have a different amount of CO a different amount of H_2 and different amount of Methanol, than we originally had. But the important thing is that we return to that ratio which is the equilibrium constant k. So lets look at a drawing here of what is going on. Here I have my reaction at equilibrium in the first panel. I go to the second panel, I have added additional carbon monoxide. Now the reaction needs to return to equilibrium. In order to do this, it needs to consume Now the reaction needs to return to equilibrium. In order to do this, it needs to consume one of our carbon monoxides and two of our hydrogens. And when it does that, it will form a molecule of CH_3OH so if we look over here after equilibrium has been re-established. We see we have increased the amount of CH_3OH present. We have gone down to having 3 carbon monoxides present, and we have also lost two of our hydrogens because they were consumed. When that carbon monoxide reacted. So we are now back at equilibrium. Though have a different number of molecules present we are going to maintain that same ratio. Now we can look at the same reaction and see what happens when we change the concentration of a product. Since there is only one product, we will be looking at methanol. If I increase the amount of methanol present. I see that the reaction now has too much methanol If I increase the amount of methanol present. I see that the reaction now has too much methanol I need to get rid of some of that methanol I see that the reaction now has too much methanol I need to get rid of some of that methanol and so as a result we are going to see the decomposition into carbon monoxide and hydrogen. So we are going to lose some of out CH_3OH in order to get back to that equilibrium ratio. So if I increase the concentration of methanol I see that I am going to favor the formation of more reactants. If I decrease the amount of CH_3OH present, if I remove some of that methanol so now looking at my equilibrium constant k equals products over reactants. I have lost some of my methanol and I need to make up for that loss. So I need to produce some more methanol to make up some of what I have lost. But in the process I am also going to decrease the amount of reactants present But in the process I am also going to decrease the amount of reactants present and as a result I will get back to product, reactant ratio decrease the amount of reactants present and as a result I will get back to product, reactant ratio that is equal to the equilibrium constant k. And so if we are decreasing the amount of products I am going to favor the formation of more product. I need to compensate for that lost amount of CH_3OH. Now these are just two examples of what Le Chatelier's principle is telling us. It is all about responding to some disturbance to the equilibrium. In these two scenarios we looked at increasing and decreasing the amounts of reactants and products. We can also look at our system at equilibrium looking at our same reaction that we looked at before now what we are looking at is we are adding more methanol so here we have now what we are looking at is we are adding more methanol so here we have added more methanol to the system and what we see happening is that when an equilibrium is re-established actually lose one of those methanol molecules and actually form more carbon monoxide and I formed some more hydrogen molecules, so these are my molecules that form And what happens, is that now I form more of my reactants I am re-establishing that equilibrium by the addition of methanol. We can look at a similar idea for the removal of methanol. We would just see that the formation of our products is favored instead of the formation of reactants that's favored when we add methanol.

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